Question: Is ${891236}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {891236}= &&{8}\cdot100000+ \\&&{9}\cdot10000+ \\&&{1}\cdot1000+ \\&&{2}\cdot100+ \\&&{3}\cdot10+ \\&&{6}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {891236}= &&{8}(99999+1)+ \\&&{9}(9999+1)+ \\&&{1}(999+1)+ \\&&{2}(99+1)+ \\&&{3}(9+1)+ \\&&{6} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {891236}= &&\gray{8\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{1\cdot999}+ \\&&\gray{2\cdot99}+ \\&&\gray{3\cdot9}+ \\&& {8}+{9}+{1}+{2}+{3}+{6} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first five terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${891236}$ is divisible by $9$ if ${ 8}+{9}+{1}+{2}+{3}+{6}$ is divisible by $9$ Add the digits of ${891236}$ $ {8}+{9}+{1}+{2}+{3}+{6} = {29} $ If ${29}$ is divisible by $9$ , then ${891236}$ must also be divisible by $9$ ${29}$ is not divisible by $9$, therefore ${891236}$ must not be divisible by $9$.